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Simple Velocity question

  • Thread starter ritwik06
  • Start date
  • #1
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Homework Statement



The instantaneous velocity of a parrticle is given by the expression:
[tex]v=u sin \omega t [/tex]
The time of complete cycle is [tex]T= \frac{2 \pi}{\omega} [/tex]

Find out the average speed!


The Attempt at a Solution


After integration;
i get the following answer:
[tex] \frac{2u} { \pi}[/tex]

tell me if I am wrong I will show my working if I am wrong.
Thanks a lot!
 

Answers and Replies

  • #2
103
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Hi ritwik

The average speed over the time of a complete cycle? I think you made a mistake. Look at the sin function again and think of it's average value. It can be done without integrating or any calculus whatsoever. Or am I wrong?
 
  • #3
580
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Hi ritwik

The average speed over the time of a complete cycle? I think you made a mistake. Look at the sin function again and think of it's average value. It can be done without integrating or any calculus whatsoever. Or am I wrong?
Lol!! Thats so clever armis. But whatever your method might be, the answer will be same. So what did u get?
Dont help people unless u get the answer urself. It might confuse the beginners like me all the more.
 
  • #4
alphysicist
Homework Helper
2,238
1
Hi ritwik06,

Your result looks right to me.
 
  • #5
ritwik, your result is wrong.
what limits did you take for integrating?
 
  • #6
103
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Lol!! Thats so clever armis
Well, I only wanted to help and solve the problem in the process. So were is no reason to laugh at me

I just thought that if velocity varies from u to -u then the average value will be just zero. Of course if we are talking about the absolute value of u then [tex] \frac{2u}{\pi} [/tex] is the answer

Dont help people unless u get the answer urself. It might confuse the beginners like me all the more.
Sorry, I didn't mean that
 
  • #7
alphysicist
Homework Helper
2,238
1
Hi jablonsky27,

ritwik, your result is wrong.
what limits did you take for integrating?
I got the same answer as ritwik. Are you integrating the correct function (they ask for the average speed)?
 
  • #8
hi alphysicist,
you dont even have to integrate the given function. the average value of a sine(or cosine) over one cycle is always zero. the velocity function mentioned, is a sine with no additional constant term. so its average value too over one cycle will be zero.
if, you really want to calculate it,


Vaverage = (1/T).[tex]\int[/tex](u.sinwt) over the limits 0 to T,

= (u/T).[-cos(w.2/[tex]\Pi[/tex]T) + cos(0)]

= (u/T).[-1 + 1]

= 0

hopefully, it makes sense. first time with Latex. :biggrin:
 
  • #9
580
0
hi alphysicist,
you dont even have to integrate the given function. the average value of a sine(or cosine) over one cycle is always zero. the velocity function mentioned, is a sine with no additional constant term. so its average value too over one cycle will be zero.
if, you really want to calculate it,


Vaverage = (1/T).[tex]\int[/tex](u.sinwt) over the limits 0 to T,

= (u/T).[-cos(w.2/[tex]\Pi[/tex]T) + cos(0)]

= (u/T).[-1 + 1]

= 0

hopefully, it makes sense. first time with Latex. :biggrin:
Well if you go back and read the question, its written in bold face:
average speed
 
  • #10
580
0
Well, I only wanted to help and solve the problem in the process. So were is no reason to laugh at me

I just thought that if velocity varies from u to -u then the average value will be just zero. Of course if we are talking about the absolute value of u then [tex] \frac{2u}{\pi} [/tex] is the answer



Sorry, I didn't mean that
Hey, I wasnt laughing at you dear. I was just wonering how u could take that out without integration. I mean the average speed. Anyways, thanks a lot for the effort.
 
  • #11
alphysicist
Homework Helper
2,238
1
jablonsky,

hi alphysicist,
you dont even have to integrate the given function. the average value of a sine(or cosine) over one cycle is always zero. the velocity function mentioned, is a sine with no additional constant term. so its average value too over one cycle will be zero.
if, you really want to calculate it,


Vaverage = (1/T).[tex]\int[/tex](u.sinwt) over the limits 0 to T,

= (u/T).[-cos(w.2/[tex]\Pi[/tex]T) + cos(0)]

= (u/T).[-1 + 1]

= 0

hopefully, it makes sense. first time with Latex. :biggrin:
I see ritwik already remarked about how we are looking for the average speed (not average velocity). So the integral to calculate would be:




[tex]
\mbox{average speed }= \frac{1}{T}\ \int\limits_0^Tdt\ |u \sin(\omega t)|
[/tex]

Or you could restrict the integral to just one-fourth of a period.
 
  • #12
oh ya. sorry. should learn to read questions more carefully in the future.
 

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